The system of equations

/[/frac{xy}{x + y} = 1, / quad / frac{xz}{x + z} = 2, / quad / frac{yz}{y + z} = 3/]has exactly one solution. What is $z$ in this solution?

z = - 12

Step-by-step explanation:

The given system of equations is:

xy / (x + y) = 1 ... (1)

xz / (x + z) = 2 ... (2)

yz / (y + z) = 3 ... (3)

From (1) : x + y = xy

=> y = xy - x

y = x (y - 1)

x = y / (y - 1) ... (4)

From (2) : 2 (x + z) = xz

=> 2x + 2z = xz

2x = xz - 2z

2x = z (x - 2)

z = 2x / (x - 2) ... (5)

From (3) : 3 (y + z) = yz

=> 3y + 3z = yz

3y = yz - 3z

3y = z (y - 3)

z = 3y / (y - 3) ... (6)

Comparing (5) and (6)

2x / (x - 2) = 3y / (y - 3)

2x (y - 3) = 3y (x - 2)

2xy - 6x = 3xy - 6y

6 (y - x) = xy ... (7)

But from (1) : xy = x + y

Using this in (7), we have

6 (y - x) = x + y

6y - y - 6x - x = 0

5y - 7x = 0

5y = 7x

x = 5y/7 ... (8)

Using this in (4)

5y/7 = y / (y - 1)

1 / (y - 1) = 5/7

(y - 1) = 7/5

y = 1 + 7/5

y = 12/5 ... (9)

Using this in (8)

x = 5 (12/5) / 7 = 12/7 ... (10)

Using (10) in (5)

z = 2x / (x - 2)

z = 2 (12/7) : (12/7 - 2)

= 24/7 : - 2/7

= 24/7 * (-7/2)

= - 24/2 = - 12

z = - 12.